Derivatives of Inverse Trigonometric Functions

In this section, we will learn the derivative formulas of the inverse trigonometric functions with their proofs.

Now we calculate the derivative of $\sin^{-1}x$ (or arc sin x).

Derivative of $\sin^{-1}x$

\[\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}} \, (|x|<1)\]

Proof: Note that $\sin^{-1}x$ is defined when $|x| \leq 1.$ We assume that
\[y=\sin^{-1}x.\] As $|x| \leq 1$, we must have that $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.$ \[y=\sin^{-1}x \Rightarrow x=\sin y\] Differentiating with respect to $y$, we have \[\frac{dx}{dy}=\frac{d}{dx}(\sin y)=\cos y\] \[\Rightarrow \frac{dx}{dy}=1-\sin^2 y\] \[\Rightarrow \frac{dx}{dy}=\sqrt{1-x^2}\] Therefore, we deduce that \[\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{\sqrt{1-x^2}}\] So the derivative of $\sin^{-1}x$ is $\frac{1}{\sqrt{1-x^2}}$, where $|x|\leq 1.$ ♣

 

Next, we find the derivative of $\cos^{-1}x$ (or arc cos x).

Derivative of $\cos^{-1}x$

\[\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}} \, (|x|<1)\]

Proof: We know that \[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\] Differentiating with respect to $x$, we get \[\frac{d}{dx}(\sin^{-1}x+\cos^{-1}x)=\frac{d}{dx}(\frac{\pi}{2})\]\[\Rightarrow \frac{d}{dx}(\sin^{-1}x)+\frac{d}{dx}(\cos^{-1}x)=0\]

[as the derivative of a constant is zero]

\[\Rightarrow \frac{1}{\sqrt{1-x^2}}+\frac{d}{dx}(\cos^{-1}x)=0\] \[\Rightarrow \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}\] So the derivative of $\cos^{-1}x$ is $-\frac{1}{\sqrt{1-x^2}}$, where $|x|\leq 1.$ ♣

 

Let us now calculate the derivative of $\tan^{-1}x$ (or arc tan x).

Derivative of $\tan^{-1}x$

\[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} \, (-\infty<x<\infty)\]

Proof: Note that $\tan^{-1}x$ is defined when $-\infty<x<\infty.$ We assume that \[y=tan^{-1}x.\] As $-\infty<x<\infty$, we must have that $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}.$ \[y=\tan^{-1}x \Rightarrow x=\tan y\] Differentiating with respect to $y$, we have \[\frac{dx}{dy}=\frac{d}{dx}(\tan y)=\sec^2 y\] \[\Rightarrow \frac{dx}{dy}=1+\tan^2 y\] \[\Rightarrow \frac{dx}{dy}=1+x^2\] Therefore, we deduce that \[\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{1+x^2}\] So the derivative of $\tan^{-1}x$ is $\frac{1}{1+x^2}$, where $-\infty<x<\infty.$ ♣

 

Next, we find the derivative of $\cot^{-1}x$ (or arc cot x).

Derivative of $\cot^{-1}x$

\[\frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1+x^2} \, (-\infty<x<\infty)\]
Proof:  We know that \[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\] Differentiating with respect to $x$, we get \[\frac{d}{dx}(\tan^{-1}x+\cot^{-1}x)=\frac{d}{dx}(\frac{\pi}{2})\] \[\Rightarrow \frac{d}{dx}(\tan^{-1}x)+\frac{d}{dx}(\cot^{-1}x)=0\]

[as the derivative of a constant is zero]

\[\Rightarrow \frac{1}{1+x^2}+\frac{d}{dx}(\cot^{-1}x)=0\] \[\Rightarrow \frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1-x^2}\] So the derivative of $\cot^{-1}x$ is $-\frac{1}{1+x^2}$, where $-\infty<x<\infty.$ ♣

 

We now calculate the derivative of $\sec^{-1}x$ (or arc sec x).

Derivative of $\sec^{-1}x$

\[\frac{d}{dx}(\sec^{-1}x)=\frac{1}{x\sqrt{x^2-1}} \, (|x|>1)\]
Proof: Note that $\sec^{-1}x$ is defined when $|x|>1.$ We assume that \[y=\sec^{-1}x.\] \[ \Rightarrow x=\sec y\] Differentiating with respect to $y$, we have \[\frac{dx}{dy}=\frac{d}{dx}(\sec y)=\sec y \tan y\] \[\Rightarrow \frac{dx}{dy}=\sec y \sqrt{\sec^2y-1}\] \[\Rightarrow \frac{dx}{dy}=x\sqrt{x^2-1}\] Therefore, we deduce that \[\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{x\sqrt{x^2-1}}\] So the derivative of $\sec^{-1}x$ is $\frac{1}{x\sqrt{x^2-1}}$, where $|x|>1.$ ♣

 

Next, we find the derivative of $\text{cosec}^{-1}x$ (or arc cosec x).

Derivative of $\text{cosec}^{-1}x$

\[\frac{d}{dx}(\text{cosec}^{-1}x)=-\frac{1}{x\sqrt{x^2-1}} \, (|x|>1)\]
Proof: We know that \[\sec^{-1}x+\text{cosec}^{-1}x=\frac{\pi}{2}\] Differentiating with respect to $x$, we get \[\frac{d}{dx}(\sec^{-1}x+\text{cosec}^{-1}x)=\frac{d}{dx}(\frac{\pi}{2})\] \[\Rightarrow \frac{d}{dx}(\sec^{-1}x)+\frac{d}{dx}(\text{cosec}^{-1}x)=0\]

 [as the derivative of a constant is zero]

\[\Rightarrow \frac{1}{x\sqrt{x^2-1}}+\frac{d}{dx}(\text{cosec}^{-1}x)=0\] \[\Rightarrow \frac{d}{dx}(\text{cosec}^{-1}x)=-\frac{1}{x\sqrt{x^2-1}}\] So the derivative of $\text{cosec}^{-1}x$ is $-\frac{1}{x\sqrt{x^2-1}}$, where $|x|> 1.$ ♣